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AY 20: Basic Astronomy and the Galaxy Solution Set 3 November 6, 2009 Swarnima Manohar [email protected] 1. C&O Problem 7.3 (a) An eclipse will just occur when the common tangent is along the line of sight, as shown in the figure. In this case, Sin90 ° i Cosi r1 x r2 ax r1 r2 a x x r1 r2 a The minimum i at which an eclipse occurs is then i ArcCos r1 r2 a (b) Printed by Mathematica for Students 2 Sol3_09.nb i ArcCos 11 Rsun 2 AU 180 . Rsun 6.96 1010 , AU 1.496 1013 i 88.5337 Thus, i = 88.5°. 2. C&O Problem 7.6 (a) mA mB vBr 22.4 4.14 5.4 vAr (b) Using Eq. 7.6 in Carroll and Ostlie, m A mB P vAr vBr 3 2G Sini3 g . P 6.31 365.25 24 3600, G 6.67 108 , vAr 5.4 105 , vBr 22.4 105 , i 2 mA mB 1.02085 1034 g which is 5.13 MŸ . (c) mA Solve 4.14, mA mB 5.13 Msun , mA, mB mA 0. 4.13195 Msun , mB 0. 0.998054 Msun mB m A = 4.13 MŸ , mB = 0.998 MŸ . (d) Assuming the orbital separation is much larger than the stellar radii, and that the orbits are circular, we can treat the velocity of the stars during eclipse as completely in the plane of the sky. For circular orbits, the maximum radial velocities given are the constant velocities throughout the orbit. The relative velocity is then in km/s v 5.4 22.4 v 27.8 It takes a time tb - ta for the smaller disk to enter the larger one, so the radius of the smaller star must be rs 1 2 v tab cm . v 27.8 105 , tab 0.58 24 3600 rs 6.96557 1010 cm which is about 1 RŸ . (Note that assuming iº 90° as we do here implies the transit is across a diameter of the large disk.) Now consider the point on the smaller disk that first eclipses the larger. From time ta to tc this point traces the diameter of the larger disk (see Fig. 7.9 in Carroll and Ostlie), so the radius of the larger disk is Printed by Mathematica for Students Sol3_09.nb rl 1 2 v tcb tab cm . v 27.8 105 , tab 0.58 24 3600 , tcb 0.64 24 3600 rl 1.46517 1011 cm which is 2.11 RŸ . (d) Following example 7.3.2 in Carroll and Ostlie, the ratio of fluxes between the primary minimum and maximum light is B p B0 = 1005.40-9.205 = 0.030, and the ratio of fluxes between the secondary minimum and maximum light is Bs B0 = 1005.40-5.445 = 0.964. Then Printed by Mathematica for Students 3 4 Sol3_09.nb Ts 1 B p B0 Tl Ts 14 1 Bs B0 . Bp B0 0.030, Bs B0 0.964 2.27833 Tl The derivation in the text assumes that the smaller star is hotter, i.e. that the primary eclipse is when the smaller star passes behind the larger. Can we back this up with the data? Assuming this is true, then in the primary eclipse we see only the larger star, which gives 100m0 -m p 5 = 1005.40-9.205 = 3.02% of the total brightness. In the secondary, we then expect to see the remaining 96.98% of the flux from the smaller star, plus the fraction rs rl 2 = 0.225 (from part c) of the big star’s 3.02% that is unob- structed, for a total of 97.7%. We observe 100m0 -ms 5 = 1005.40-5.445 = 96.38%, which is close considering we only two significant digits in the times. If we instead imagine that the larger star is hotter, so that the primary eclipse occurs when the smaller star is in front, we find that the 96.38% of light visible in the secondary is due to the larger star, and that during the primary we expect 0.225 of this (plus the 3.62% contribution of the smaller star), which is far in excess of the 3.02% observed. Another way to estimate the temperature ratio is to note that the primary minimum is 0.030 of the maximum brightness, so that 0.030 T4s Rs2 T4l Rl2 T4l Rl2 Ts 1 1 0.030 Tl Rl 2 14 3.5 Rs using the radii from (d). This does not agree particularly well with the 2.3 we found above, indicating that the timing data in this problem are not consistent with the photometry under the assumption that the smaller star passes directly across a diameter of the larger star’s disk. For this homework either answer is acceptable. 3. C&O Problem 7.11 For simplicity, assume circular orbits. (In any case, eccentricity can be deduced from distortions of the radial velocity curve from sinusoidal.) Denoting the masses of the planet and star by M p and Ms , Kepler’s third law says P2 4 2 G Mp Ms a3 4 2 a3 G Ms This is because M p ` Ms , so a G Ms P2 13 4 2 The orbital velocity of the planet is approximately vp 2a P 2 G Ms 13 P and that of the star is reduced approximately by the mass ratio: vs Mp vp Ms Printed by Mathematica for Students Sol3_09.nb 5 We can measure the period P and estimate Ms by determining its spectral type with our spectrum. Thus, we know v p . If we could measure vs , then the above equations would allow us to solve for M p . However, what we measure is vrs = vs sin i, the radial lower limit on M p = Ms vs v p , but cannot say more. component of the star’s velocity, and the orbital inclination i is in general unknown. Since vrs is a lower limit on vs , we get a 4. C&O Problem 7.13 Under the assumptions of a uniform solar disk and no luminosity from Jupiter, the decrease is just given by the fraction of the solar disk blocked by Jupiter: RJ2 Rü2 RJ 7.1 109 cm 2 7 1010 cm Rü 2 1.1 5. C&O Problem 9.20 For a plane grey atmosphere in LTE under the Eddington approximation, Carroll and Ostlie derive 4 3 I Frad 2 3 in Eqn 9.50. Using Frad = sT4e (by definition) and I = S = B = sT4 /p (in LTE), this becomes T4 3 4 T4e 2 3 From this we see that T = Te at t = 2/3. (Note that along any line of sight, you are seeing photons from an average t = 1; however, when averaged over angles, this corresponds to an average vertical depth of tv =2/3.) This can also be derived by assuming a liinear source function S = a +bt, which is another form of the Eddington approximation. 6. C&O Problem 10.11 The goal is to approximate the expression (eqn 10.62) 1 44.027 T8 3 50.9 2 Y3 T3 8 f3 as a power law of the form: 3 T8 near T = 108 T8 = 1. We equate the two equations, take log of both sides, require their log-derivatives to match, and plug in T8 = 1 to solve for a: ln T8 44.027 T8 ln T8 const ln T3 8 3 1 44.027 41 T8 7. C&O Problem 10.12 Printed by Mathematica for Students 6 Sol3_09.nb The mass of the proton 1 1 H 1.0073 u, and that of 4 2 He is 4.0026 u. In the text, the masses of 2.0141 u and 3.0160 u, but note that the mass quoted here for 2 1 2 1 H and 3 2 He are also given as H seems to + include the mass of the positron e , based on comparison with other references. To compute the Q value we subtract the total mass of the products from that of the reactants (with heat released then corresponding to Q > 0): Reaction Q 10.37 2 1.0073 2.0141 u 0.0005 u 0.5 MeV 10.38 2.0141 1.0073 3.0160 u 0.0054 u 5.0 MeV 10.39 2 3 : 0160 4.0026 2 1.0073 u 0.0148 u 13.8 MeV The conversion from mass (amu) to energy (MeV) is just multiplying by c2 . To compute the net Q per 4 2 He produced, note that the reactions (10.37) and (10.38) must each proceed twice for every production of helium in (10.39). Also, the two positrons thus produced in (10.37) will immediately annihilate with two electrons and produce 4×511 keV = 2:0 MeV. The net Q is thus 2 × 0.5 + 2× 5.0 + 13.8 + 2:0 = 26.8 MeV per 42 He. Printed by Mathematica for Students